[luogu4868]Preprefix sum

https://www.luogu.org/problemnew/show/P4868

题目大意

单点修改，查询前缀前缀和。

分析

遇到了单点修改，前缀和，很明显是要树状数组维护解决问题。
请看以下我的数列的转换
\[s1+s2+s3+...+sn\]
\[a1+a1+a2+a1+a2+a3+...+an\]
\[a1*n+a2*(n-1)+a3*(n-2)+...an*1\]
\[(a1+a2+a3+...+an)*N - (a2+a3^2+a4^3+...+an^{n-1})\]

ac代码

#include <bits/stdc++.h>
#define ll long long
#define ms(a, b) memset(a, b, sizeof(a))
#define inf 0x3f3f3f3f
#define N 100005
using namespace std;
template <typename T>
inline void read(T &x) {
    x = 0; T fl = 1;
    char ch = 0;
    while (ch < '0' || ch > '9') {
        if (ch == '-') fl = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    x *= fl;
}
struct bittree {
    #define lowbit(x) (x&(-x))
    ll tr[N];
    int n;
    void add(int k, ll val) {
        for (int i = k; i <= n; i += lowbit(i)) 
            tr[i] += val;
    }
    ll query(int x) {
        ll res = 0;
        for (int i = x; i; i -= lowbit(i)) 
            res += tr[i];
        return res;
    }
}tr1, tr2;
int n, m;
ll a[N];
int main() {
    read(n); read(m);
    tr1.n = tr2.n = n;
    for (int i = 1; i <= n; i ++) {
        read(a[i]);
        tr1.add(i, a[i]);
        tr2.add(i, a[i] * (i - 1));
    }
    while (m --) {
        char opt[10];
        scanf("%s", opt);
        ll x, y;
        if (opt[0] == 'Q') {
            read(x);
            printf("%lld\n", (ll)(tr1.query(x) * x) - 1ll * tr2.query(x));
        }
        else {
            read(x); read(y);
            tr1.add(x, y - a[x]);
            tr2.add(x, (y - a[x]) * (x - 1));
            a[x] = y;
        }
    }
    return 0;
}